【PAT甲級】Hello World for U
- 2019 年 11 月 8 日
- 筆記
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本文鏈接:https://blog.csdn.net/weixin_42449444/article/details/89435176
Problem Description:
Given any string of N (≥5) characters, you are asked to form the characters into the shape of U
. For example, helloworld
can be printed as:
h d e l l r lowo
That is, the characters must be printed in the original order, starting top-down from the left vertical line with n1 characters, then left to right along the bottom line with n2 characters, and finally bottom-up along the vertical line with n3 characters. And more, we would like U
to be as squared as possible — that is, it must be satisfied that n1=n3=max { k | k≤n2 for all 3≤n2≤N } with n1+n2+n3−2=N.
Input Specification:
Each input file contains one test case. Each case contains one string with no less than 5 and no more than 80 characters in a line. The string contains no white space.
Output Specification:
For each test case, print the input string in the shape of U as specified in the description.
Sample Input:
helloworld!
Sample Output:
h ! e d l l lowor
解題思路:
膜柳神~ 真的大佬(博文鏈接:https://www.liuchuo.net/archives/2053),她的思路是:假設n = 字元串長度 + 2,因為2 * n1 + n2 = n,且要保證n2 >= n1, n1儘可能地大,分類討論:① 如果n % 3 == 0,n正好被3整除,直接n1 == n2 == n3; ② 如果n % 3 == 1,因為n2要比n1大,所以把多出來的那1個給n2;③ 如果n % 3 == 2, 就把多出來的那2個給n2;所以得到公式:n1 = n / 3,n2 = n / 3 + n % 3。初始化char型數組為空格,然後將字元串按照u型填充進去,最後輸出這個數組u即可。
AC程式碼:
#include <bits/stdc++.h> using namespace std; int main() { string str; getline(cin,str); int n = str.length() + 2; int n1 = n/3; //n1是左右兩條豎線從上到下的字元個數 int n2 = n1 + n%3; //n2是底部橫線從左到右的字元個數 char arr[30][30]; memset(arr,' ',sizeof(arr)); int cnt = 0; //用來記錄當前字元的下標 for(int i = 0; i < n1; i++) //先排列左邊那條豎線的n1個字元 hell { arr[i][0] = str[cnt++]; } for(int i = 1; i < n2-1; i++) //再排列底部那條橫線(除去最左最右字元外)的那n2-2個字元 owo { arr[n1-1][i] = str[cnt++]; } for(int i = n1-1; i >= 0; i--) //最後排列右邊那條豎線的n1個字元 rld! { arr[i][n2-1] = str[cnt++]; } //列印U型圖案 for(int i = 0; i < n1; i++) { for (int j = 0; j < n2; j++) { cout << arr[i][j]; } cout << endl; } return 0; }